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NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement

Edited By Vishal kumar | Updated on Mar 22, 2025 12:04 AM IST

Chapter 1 - Units and Measurement deals with the measurement of different physical quantities, for example, w hen we check the temperature of water with a thermometer, we use units like Celsius to quantify the temperature. The exercises include 17 questions, numbered from 1.1 to 1.17.

For example, when you measure the length of a table with a ruler, you are using the concept of "measurement" and "units" from this chapter. A ruler gives the length in centimeters (cm) or inches.

In the CBSE board exams, a significant portion of questions are derived directly from the NCERT concepts and exercises. This is where tools like NCERT Solutions for Physics Class 11 Chapter 1, "Units and Measurement," become incredibly valuable. You’ll find detailed solutions for each question, along with important formulas from the chapter to help you prepare for class tests, assignments, homework, and exams.

This Story also Contains
  1. NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement: Download PDF
  2. Important topics in Chapter 1, "Units and Measurements"
  3. Important Tips to Prepare Well Unit and Measurement Chapter
  4. NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement:
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement: Download PDF

It covers the essential topics and tools that you’ll need to understand and solve the problems in Chapter 1, which will help you develop a strong foundation in units, measurement, and related concepts

Download PDF

Access the class 11 physics chapter 2 exercise solutions unit and measurement

Q1.1 Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to ..... m3

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ... (mm)2

(c) A vehicle moving with a speed of 18 kmh1 covers....m in 1 s

(d) The relative density of lead is 11.3. Its density is .... gcm3 or .... kgm3 .

Answer:

(a) We know, 1cm=.01m (Tip: Divide by 100 to convert cm to m)

The volume of a cube of side a = a3 m3

Volume of cube of side 1 cm (i.e, .01 m ) = (.01)3=106 m3

(b) We know, 1cm=10mm (Tip: Multiply by 10 to convert cm to mm)

The surface area of a solid cylinder of radius r and height h = 2(πr2)+2πrh=2πr(r+h)

Required area = 2π(20)( 20+100)mm3=40π(120)mm3=4800π mm3=1.5×104 mm3

(c) (Tip: multiply by 5/18 to convert kmh1 to ms1 )

Ditance covered = Speed×time=(18×5/18)ms1×(1s)=5m

(d) Density = Relative Density × Density of water

(Density of water = 1 gcm3=1000 kgm3 )

Density of lead=

(11.3)×1 gcm3=11.3 gcm3(11.3)×1000 kgm3=11300 kgm3=1.13×104 kgm3

Q 1.2 : Fill in the blanks by suitable conversion of units

(a) 1kgm2s2=        gcm2s2

b) 1m=    ly

(c) 3.0 ms2=    kmh2

(d) G=6.67×1011Nm2(kg)2=     (cm)3s2g1

Answer:

(a) (1kg1000g; 1m100cm)

1kgm2s2=(1000g)(100cm)2s2=(103×104)gcm2s2=107 gcm2s2

(b) 1m=1.057×1016ly (1 ly = Distance travelled by light in 1 year )

(c) (1m103km; (60×60=)3600s1hr)

3.0ms2=3.0(103km)(1/3600hr)2=3.0×(3600)2/1000 kmhr2=3.9×104 kmhr2

(d) (NJ.s=kgms2; 1m100cm; 1kg1000g)

G=6.67×1011(kgms2)m2kg2=(6.67×1011)(m)3s2kg1=6.67×108cm3s2g1

Q 1.3 : A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J=1kgm2s2 . Suppose we employ a system of units in which the unit of mass equals α kg , the unit of length equals β m , the unit of time is γ s . Show that a calorie has a magnitude 4.2 α1β2γ2 in terms of the new units.

Answer:

Given,

1Cal=4.2(kg)(m)2(s)2

Given new unit of mass = αkg (In old unit 1kg corresponded to a unit mass, but in new unit αkg corresponds to a unit mass)

In terms of the new unit, 1kg=1/α=α1

Similarly in terms of new units , 1m=1/β=β1 and 1s=1/γ=γ1

1Cal=4.2(kg)(m)2(s)2=4.2(α1)(β1)2(γ1)2=4.2α1β2γ2

Topic

Q 1.4
: Explain this statement clearly: “To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary :

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light

Answer:

The given statement is true. A dimensional quantity may be small with respect to one reference and maybe large with respect to another reference. Hence, we require a standard reference to judge for comparison.

(a) An atom is a very small object with respect to a tennis ball. (but larger than an electron!)

(b) A jet plane moves with great speed with respect to a train.

(c) The mass of Jupiter is very large as compared to an apple.

(d) The air inside this room contains a large number of molecules as compared to in your lungs.

(e) A proton is much more massive than an electron

(f) The speed of sound is less than the speed of light

Q 1.5 : A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Answer:

Distance between Sun and Earth = Speed of light x Time taken by light to cover the distance

Speed of light = 1 unit

Time taken by light to reach earth is 8 minute 20 seconds

Time taken = (8×60)+20=500s

The distance between Sun and Earth = 1 x 500 = 500 units.

Q 1.6 : Which of the following is the most precise device for measuring length :

(a) a vernier callipers with 20 divisions on the sliding scale

(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale

(c) an optical instrument that can measure length to within a wavelength of light?

Answer:

To judge which tool is more precise, we have to find out their least count. Least count defines the margin of error and hence the precision. Hence the instrument with lower least count will be more precise.

(a) Least count = 1MSD1VSD=1(19/20)=1/20=0.05 cm (Taking 1 MSD as 1 mm)

(b) Least count = pitch/ number of divisions

= 1/10000=0.001cm

(c) least count = wavelength of light = 400nm to 700nm, that is in the range of 107 m

Therefore, the optical instrument is the most precise device used to measure length.

Q 1.7 : A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?

Answer:

Given,

Magnification of Microscope = 100

The average width of hair under the microscope = 3.5 mm

(20 observations were made to calculate the average i.e. 3.5 mm as an experimental procedure. No need in our calculations.)

(Note: When magnified, the width is 3.5 mm. Hence actual width will be less by a factor of magnification value)

The average thickness of hair = 3.5mm100=0.035 mm.

Q 1.8 Answer the following :

(a)You are given a thread and a metre scale. How will you estimate the diameter of the thread?

(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Answer:

(a) Take the thread and wrap it around the metre scale. Make sure the coils are packed closely without any space in between. If the diameter of the thread is d and number of turns obtained are n, then (n x d) corresponds to the marking on the metre scale, l.

Therefore, the diameter of the thread would be, d = l/n

(b) Theoretically, by increasing the number of divisions on the circular divisions, the value of least count decreases and hence accuracy increases.(Lower the value of least count, better will be the reading)

But practically, the number of divisions can be increased only up to a certain limit. (Also two adjacent divisions cannot be separated by a distance less than the human eye resolution!)

(c) With an increase in the number of observations, the accuracy of the experiment increases as the error is now distributed over a large range. Hence, a set of 100 measurements of the diameter is expected to yield a more reliable estimate than a set of 5 measurements.

Q 1.9 : The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2 . What is the linear magnification of the projector-screen arrangement?

Answer:

Given,

Area of the house in the photo = 1.75 cm2

Area of the house on the screen = 1.55m2=1.55×104cm2

Arial magnification, ma = Area on the screen / area on photo = 1.55×1041.75=0.886×104

Linear magnification of the projector- screen arrangement ml=ma=.886×104=94.1

Q 1.10 : State the number of significant figures in the following :
(a)0.007 m2

(b) 2.64×1024kg

(c) 0.2370 gcm3

(d) 6.320 J

(e) 6.032 Nm2

(f) 0.0006032 m2

Answer:

(a) The given value is 0.007 m2 .

Since, the number is less than 1 , the zeros on the right to the decimal before the first non-zero integer is insignificant. So, the number 7 is the only significant digit.

It has 1 significant digit.

(b) The value is 2.64×1024kg

For the determination of significant values, we do not consider the power of 10 (Number is not less than 1). The digits 2, 6, and 4 are significant figures.

It has 3 significant digits.

(c)The value is 0.2370 gcm3 .

For the given value with decimals, all the numbers 2, 3, 7 and 0 are significant.

It has 4 significant digits.

(d) The value is 6.320 J .

It has 4 significant digits.

(e) The value is 6.032 Nm2 .

All the four digits are significant as the zeros in between two non-zero values are also significant.

It has 4 significant digits.

(f) The value is 0.0006032 m2

Same as (a), first three zeroes after the decimal is insignificant. Only 6, 0, 3, 2 are significant.

It has 4 significant digits.

Q 1.11 : The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Answer:

Given,

Length, l = 4.234 m ; Breadth, b = 1.005 m; Height, h = 2.01 cm = 0.0201 m

The length has 4 significant figures

The breadth has 4 significant figures

The height has 3 significant figures (Since the number is less than 1, hence zeroes after decimal before the first non-zero integer is insignificant)

We know,

Surface area of a cuboid = 2(l x b + b x h + h x l)

= 2(4.234×1.005+1.005×0.0201+0.0201×4.234)

= 2(4.255+0.020+0.085)

= 8.72 m2 (Note: For addition/subtraction , the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal. )

Volume = l x b x h

= 4.234×1.005×0.0201 = 0.0855 m3 ( 3 significant digits)

(Note: For Multiplication The LEAST number of significant digits in any number determines the number of significant figures in the answer; i.e decided by the number of significant digits )

The area has three significant values 2, 7 and 8.

The volume has three significant values 5, 5 and 8.

Q 1.12 : The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Answer:

Given,

The mass of the box = 2.30 kg

and the mass of the first gold piece = 20.15 g = 0.02015 kg

The mass of the second gold piece = 20.17 g = 0.02017 kg

(a) The total mass = 2.30 + 0.02015 + 0.02017 = 2.34032 kg

Since one is the least number of decimal places, the total mass = 2.3 kg.

(Note: For addition/subtraction , the number of places after the decimal point in the answer is less than or equal to the number of decimal places in every term in the sum; i.e decided by the numbers after the decimal. )

Q 1.13 : A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c . (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes : m=mo(1v2)1/2

Guess where to put the missing c.

Answer:

The relation given is m=mo(1v2)1/2

Divide both sides by mo ; L.H.S becomes m/mo which is dimensionless.

Hence, R.H.S must be dimensionless too. (After Dividing by mo !)

1(1v2)1/2 can be dimensionless only when v(v/c)

Therefore, the dimensional equation is m=mo(1(vc)2)1/2

Q 1.14 : The unit of length convenient on the atomic scale is known as an angstrom and is denoted by: 1Å = 10^{-10} m$ . The size of a hydrogen atom is about 0.5Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?

Answer:

Radius of an Hydrogen atom = 0.5 Å = 0.5 x 10 -10 m

Volume = 43πr3

= 4/3×22/7×(0.5×1010)3

= 0.524×1030m3

1 hydrogen mole contains 6.023×1023 hydrogen atoms.

The volume of 1 mole of hydrogen atom = 6.023×1023×0.524×1030

= 3.16×107m33×107m3 .

Q1.15 : One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Answer:

Radius of hydrogen atom = 0.5 Å = 0.5 x 10 -10 m (Size here refers to Diameter!)

Volume occupied by the hydrogen atom= 43πr3

= 4/3×22/7×(0.5×1010)3

= 0.524×1030m3

1 mole of hydrogen contains 6.023 x 10 23 hydrogen atoms.

Volume of 1 mole of hydrogen atom = 6.023 x 10 23 x 0.524 x 10 -30

= 3.16 x 10 -7 m 3

Vm=22.4L=22.4×103m3

VmVa=22.4×1033.16×107=7.09×104

The molar volume is 7.09×104 times greater than the atomic volume.

Hence, intermolecular separation in gas is much larger than the size of a molecule.

Q 1.16 Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer:

Our eyes detect angular velocity, not absolute velocity. An object far away makes a lesser angle than an object which is close. That's why the moon (which is so far away!) does not seem to move at all angularly and thus seems to follow you while driving.

In other words, while in a moving train, or for that matter in any moving vehicle, a nearby object moves in the opposite direction while the distant object moves in the same direction. !

Q 1.17: The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107K , and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: mass of the Sun = 2.0×1030kg , radius of the Sun = 7.0×108m

Answer:

Given,

Mass of the Sun, m = 3×1030 kg

The radius of the Sun, r = 8×108 m

Volume V = 43πr3

= 43×227×(8×108)3=2145.52×1024m3

Density = Mass/Volume

= 3×10302145.52×1024=1.39×103 kgm3

Therefore, the density of the sun is in the range of solids and liquids and not gases. This high density arises due to inward gravitational attraction on outer layers due to inner layers of the Sun. (Imagine layers and layers of gases stacking up like a pile!)

Chapter 1: Units and Measurements in NCERT Class 11 Physics: Essential Formulas, and eBook Link

Boost your understanding of NCERT Class 11 Physics Chapter 1, 'Units and Measurements,' with our collection of key formulas, diagrams, and an eBook link. This resource provides you with the essential tools to master the chapter's concepts. Access the eBook for a complete and thorough learning experience.

  • Unit and Dimensions of Some Physical Quantities:


Physical Quantity

UnitDimensions
Densitykg/m³M L⁻³
ForceNewton (N)M L T⁻²
Work / EnergyJoule (J = N·m)M L² T⁻²
PowerWatt (W = J/s)M L² T⁻³
Momentumkg·m/sM L T⁻¹
Gravitational Constant (G)N·m²/kg²M⁻¹ L³ T⁻²
Angular Velocityrad/sT⁻¹
Angular Accelerationrad/s²T⁻²
Angular Momentumkg·m²/sM L² T⁻¹
Angular Frequencyrad/sT⁻¹
Moment of Inertiakg·m²M L²
TorqueN·mM L² T⁻²
FrequencyHertz (Hz)T⁻¹
Time PeriodSecond (s)T
Surface TensionN/mM T⁻²
Specific Heat CapacityJ/(kg·K)L² T⁻² Θ⁻¹
Heat / EnergyJoule (J)M L² T⁻²
Electric Dipole MomentC·mI T L
Electric FieldV/m or N/CM L T⁻³ I⁻¹
Electric Potential (Voltage)Volt (V = J/C)M L² T⁻³ I⁻¹
Electric FluxV·mM L³ T⁻³ I⁻¹
CapacitanceFarad (F = C/V)M⁻¹ L⁻² T⁴ I²
Electromotive Force (EMF)Volt (V)M L² T⁻³ I⁻¹
Magnetic FieldTesla (T)M T⁻² I⁻¹
Magnetic FluxWeber (Wb = T·m²)M L² T⁻² I⁻¹

These are the physical quantity and dimension. you can also access the chapter-wise formula for all chapters by clicking on the below-given link.


Download Ebook - Formula Sheet for Physics Class 11: Chapterwise Important Formulas With Examples, Graphs, And Points

Important topics in Chapter 1, "Units and Measurements"

  • Fundamental and Derived Quantities: Fundamental quantities are the basic building blocks of measurement, like length, mass, and time. Derived quantities are made by combining these basics, such as velocity (distance/time) or force (mass × acceleration).

  • Units and Systems: Different systems of units are used to measure physical quantities. The most commonly used system is the International System of Units (SI), which includes units like meters for length and kilograms for mass. It's the global standard for scientific measurements.

  • Significant Figures: Significant figures are the digits in a measurement that you know for sure, along with one digit that’s estimated. They show how precise and accurate your measurement is, which is crucial when working with scientific data.

  • Errors and Their Types: Errors refer to the differences between a measured value and the true or accepted value. There are two main types: systematic errors, which affect your measurements in a consistent way (like a miscalibrated instrument), and random errors, which happen unpredictably and vary from one measurement to the next.

  • Rounding Off: When reporting measurements, rounding off involves dropping extra digits that don’t affect the accuracy of your result. It helps in making the numbers easier to understand without losing precision.

  • Dimensions of Physical Quantities : Dimensions represent the fundamental units that describe a physical quantity. They help in understanding the relationship between different physical quantities.

  • Unit Conversion : Converting between different units ensures consistency and coherence in measurements, especially when working with various unit systems.
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Each of these topics plays a crucial role in understanding the concepts of units and measurements, and mastering them is essential for solving the exercise questions effectively.

Important Tips to Prepare Well Unit and Measurement Chapter

When studying the Class 11 Physics Chapter 1 exercise solutions (now Chapter 1 as per the CBSE Syllabus 2025-26), keep the following tips in mind

  • Understand the Difference Between Fundamental and Derived Quantities : Fundamental quantities are independent (like mass, length, and time), while derived quantities depend on them (such as velocity and force).
  • Get Comfortable with Different Unit Systems, Especially SI : Focus on the SI system, which is the standard for most physical measurements. Learn its basic units and their usage.
  • Practice Significant Figures and Rounding Off : Master the concept of significant figures and apply rounding off to ensure your results are precise and correctly reported.
  • Know the Types of Errors and How to Minimize Them : Differentiate between systematic and random errors and learn ways to reduce their impact on your measurements.
  • Master Unit Conversion : Learn how to convert between various units, especially across different systems like SI and CGS, to ensure your calculations remain consistent.

Also read :

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurement:

The topic and subtopic of Physics Class 11 Chapter 1 exercise solutions Units and Measurement are listed below.

Section NameTopic Name
1

Units and Measurements

1.1Introduction
1.2

The international system of units

1.3

Measurement of length

1.4

Measurement of mass

1.5

Measurement of time

1.6

Accuracy, precision of instruments and errors in measurement

1.7

Significant figures

1.8

Dimensions of physical quantities

1.9

Dimensional formulae and dimensional equations

1.1

Dimensional analysis and its applications



In Class 11 Physics Chapter 1, "Units and Measurement," measurements always need to be accompanied by a unit; without a unit, they hold no meaning. For example, when a person says the distance between his home and the nearest city is "5," this statement is meaningless unless a unit is specified, such as 5 meters, 5 centimetres, or 5 kilometres.

The NCERT Solutions for Physics Class 11 Chapter 1 begins with questions on unit conversion. These solutions provide a clear understanding of concepts like accuracy, precision, and error, which are crucial for mastering the chapter. The chapter also covers important topics such as the concepts of errors, methods for measuring errors, dimensional analysis, and significant figures. All of these topics are addressed in detail through the Class 11 Physics Chapter 1 exercise solutions, helping students grasp key concepts and solve related questions effectively.

Also, Check NCERT Books and NCERT Syllabus here

NCERT solutions for class 11 physics chapter-wise


NCERT solutions for class 11 Subject-wise


Frequently Asked Questions (FAQs)

1. How do you convert one unit to another in physics?

Multiply the given value by a conversion factor that relates the two units.

2. What are the seven fundamental units in SI system?

The seven fundamental SI units are:

  1. Meter (m) – length
  2. Kilogram (kg) – mass
  3. Second (s) – time
  4. Ampere (A) – electric current
  5. Kelvin (K) – temperature
  6. Mole (mol) – amount of substance
  7. Candela (cd) – luminous intensity
3. Why is dimensional analysis important in physics?

Dimensional analysis is important in physics because it helps check the correctness of equations, convert units, and understand relationships between physical quantities.

4. What are the important topics in Units and Measurements ?

Here are the important topics in Units and Measurements:

  1. Physical quantities
  2. Fundamental and derived units
  3. SI units
  4. Unit conversions
  5. Dimensional analysis
  6. Accuracy, precision, and errors

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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